Question

Solve the following equation and also check your result:

\[\frac{(45 - 2x)}{15} - \frac{(4x + 10)}{5} = \frac{(15 - 14x)}{9}\]

Answer 1

\[\frac{45 - 2x}{15} - \frac{4x + 10}{5} = \frac{15 - 14x}{9}\]
\[\text{ or }\frac{45 - 2x - 12x - 30}{15} = \frac{15 - 14x}{9}\]
\[\text{ or }\frac{15 - 14x}{5} = \frac{15 - 14x}{3} [\text{ Multiplying both sides by }3]\]
\[\text{ or }45 - 42x = 75 - 70x [\text{ After cross multiplication }]\]
\[\text{ or }70x - 42x = 75 - 45\]
\[\text{ or }28x = 30\]
\[\text{ or }x = \frac{30}{28}\]
\[\text{ or }x = \frac{15}{14}\]
\[\text{ Thus, } x = \frac{15}{14}\text{ is the solution of the given equation }. \]
\[\text{ Check: }\]
\[\text{ Substituting }x = \frac{15}{14}\text{ in the given equation, we get: }\]
\[\text{ L . H . S .} = \frac{45 - 2 \times \frac{15}{14}}{15} - \frac{4 \times \frac{15}{14} + 10}{5} = \frac{45 \times 7 - 15}{105} - \frac{30 + 70}{35} = \frac{300}{105} - \frac{100}{35} = 0\]
\[\text{ R . H . S .} = \frac{15 - 14 \times \frac{15}{14}}{9} = 0\]
\[ \therefore \text{ L . H . S . = R . H . S . for }x = \frac{15}{14}\]